1.) Rn-222 decays from 400 grams to 6.25 grams in 240 minutes. How long is one “half-life.

MA seconds class lever always more than 1

Hoochie [10]

It is highly helpful to know that the mechanical advantage (M.A.) of Class two levers is usually greater than one. It is because the overall length of the Effort Arm is higher than the overall length of Load Arm. It is easily known by MA is effort arm/load arm.

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The mechanical advantage of a lever of the second order is always greater than one because its effort arm is always longer than the load arm i.e. Effort arm > Load arm.

Second class lever has mechanical advantage always more than one as load is in between fulcrum and effort making the effort arm longer than the load arm.

First Class Lever -- the effort and the load on either side of the fulcrum. Some examples would be a crowbar or a seesaw. The effort is only less than the load if the load is closer to the fulcrum. The lever then acts as a force magnifier and the mechanical advantage is greater than one.

8 0

3 years ago

An inductor carries a current of 1.6 a, and the magnetic energy stored in the inductor is 0.0020 j. what is the inductance

faltersainse [42]

Energy stored= Li^2/2
Therefore L=2(E.S)/i^2
= 0.0013

8 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

(I) A car slows down from 28 m????s to rest in a distance of 88 m. What was its acceleration, assumed constant?

Nataly [62]

Answer:

The value is $a = - 4.45 m/s^2$

Explanation:

From the question we are told that

The initial speed is $u = 28 \ m/s$ at a distance of $s_1 = 0 \ m$

The final speed is $v = 0 \ m/s$ at a distance of $s_2 = 88 \ m$

Generally from the kinematic equation we have that

$v^2 = u^2 +2as$

=> $a = \frac{v^2 - u^2 }{ 2(s_2 - s_1 )}$

=> $a = \frac{0 - 28^2 }{ 2(88 - 0 )}$

=> $a = - 4.45 m/s^2$

The negative sign shows that it is decelerating

6 0

4 years ago

A nonconducting ring with a radius of 11.5 cm is uniformly charged with a total positive charge of 10.0 µC. The ring rotates at

zhuklara [117]

Answer:

$B=1.21*10^{-10}T$

Explanation:

The magnitude of the magnetic field on the axis of the ring is given by:

$B=\frac{\mu_0 IR^2}{2(r^2+R^2)^{\frac{3}{2}}}(1)$

$\mu_0$ is the permeability of free space, $I$ is the flowing current through the ring, $R$ is the ring's radius and $r$ is the distance to the center of the ring.

The flowing current through the ring is defined as the ring's charge divided into the time taken by the charge to complete one revolution, that is, the period $T=\frac{2\pi}{\omega}$ . So, we have:

$I=\frac{q}{T}\\I=\frac{q}{\frac{2\pi}{\omega}}\\\\I=\frac{\omega q}{2\pi}\\I=\frac{18\frac{rad}{s}(10*10^{-6}C)}{2\pi}\\I=2.87*10^{-5}A$

Now, replacing in (1):

$B=\frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(2.87*10^{-5}A)(0.115m)^2}{2((0.05m)^2+(0.115m)^2)^{\frac{3}{2}}}\\B=1.21*10^{-10}T$

6 0

4 years ago